Find $\lim_{x\to 4}\dfrac{x-4}{3-\sqrt{x+5}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-6$ (Choice B) B $-\dfrac{4}{3}$ (Choice C) C $-\dfrac{10}{3}$ (Choice D) D The limit doesn't exist
Substituting $x=4$ into $\dfrac{x-4}{3-\sqrt{x+5}}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x-4}{3-\sqrt{x+5}} \\\\ &=\dfrac{x-4}{3-\sqrt{x+5}}\cdot\dfrac{3+\sqrt{x+5}}{3+\sqrt{x+5}} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x-4)(3+\sqrt{x+5})}{3^2-(x+5)} \\\\ &=\dfrac{\cancel{(x-4)}(3+\sqrt{x+5})}{-1\cancel{(x-4)}} \gray{\text{Cancel out common factors}} \\\\ &={-3-\sqrt{x+5}} \text{, for }x\neq 4 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $4$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x-4}{3-\sqrt{x+5}}=-3-\sqrt{x+5}$ for all $x$ -values in the interval $(3.5,4.5)$ except for $x=4$. Therefore, $\lim_{x\to 4}\dfrac{x-4}{3-\sqrt{x+5}}=\lim_{x\to 4}-3-\sqrt{x+5}=-6$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 4}\dfrac{x-4}{3-\sqrt{x+5}}=-6$.